1. (2026·海南万宁期末)[阅读材料]小明同学发现一个规律:两个共顶点且顶角相等的等腰三角形,底角顶点连起来,在相对位置变化的同时,始终存在一对全等三角形,小明把具有这种规律的图形称为“手拉手模型”.
[材料理解](1)如图(1),△ABC与△ADE都是等腰三角形,AB = AC,AD = AE,且∠BAC=∠DAE,则有△ABD≌
[深入研究](2)如图(2),△ABC与△ADE都是等腰三角形,AB = AC,AD = AE,且∠BAC=∠DAE=90°.
①请问△ABD与△ACE是全等三角形吗?如果是全等三角形,请说明理由.
②请判断线段BD和CE的数量关系和位置关系,并说明理由.
[材料理解](1)如图(1),△ABC与△ADE都是等腰三角形,AB = AC,AD = AE,且∠BAC=∠DAE,则有△ABD≌
△ACE
;线段BD和CE的数量关系是BD=CE
.[深入研究](2)如图(2),△ABC与△ADE都是等腰三角形,AB = AC,AD = AE,且∠BAC=∠DAE=90°.
①请问△ABD与△ACE是全等三角形吗?如果是全等三角形,请说明理由.
②请判断线段BD和CE的数量关系和位置关系,并说明理由.
答案：
1.(1)$△ ACE$ $BD=CE$
(2)①$△ ABD$与$△ ACE$是全等三角形.理由如下：
$\because ∠ BAC = ∠ DAE = 90°, ∠ BAC + ∠ BAE = ∠ CAE,$
$∠ DAE + ∠ BAE = ∠ BAD,$
$\therefore ∠ CAE = ∠ BAD.$
在$△ ABD$和$△ ACE$中,
$\begin{cases} AB=AC, \\ ∠ BAD=∠ CAE, \\ AD=AE, \end{cases}$
$\therefore △ ABD ≌ △ ACE (\mathrm{SAS}).$
②$BD=CE, BD ⊥ CE$.理由如下：
如图,设$AB$,$CE$相交于点$O$.
$\because △ ABD ≌ △ ACE,$
$\therefore BD=CE, ∠ ABD = ∠ ACE.$
$\because ∠ BPC + ∠ ABD + ∠ BOE = ∠ BAC + ∠ ACE + ∠ AOC,$
$∠ BOE = ∠ AOC,$
$\therefore ∠ BPC = ∠ BAC = 90°,$
$\therefore BD ⊥ CE.$
1.(1)$△ ACE$ $BD=CE$
(2)①$△ ABD$与$△ ACE$是全等三角形.理由如下：
$\because ∠ BAC = ∠ DAE = 90°, ∠ BAC + ∠ BAE = ∠ CAE,$
$∠ DAE + ∠ BAE = ∠ BAD,$
$\therefore ∠ CAE = ∠ BAD.$
在$△ ABD$和$△ ACE$中,
$\begin{cases} AB=AC, \\ ∠ BAD=∠ CAE, \\ AD=AE, \end{cases}$
$\therefore △ ABD ≌ △ ACE (\mathrm{SAS}).$
②$BD=CE, BD ⊥ CE$.理由如下：
如图,设$AB$,$CE$相交于点$O$.
$\because △ ABD ≌ △ ACE,$
$\therefore BD=CE, ∠ ABD = ∠ ACE.$
$\because ∠ BPC + ∠ ABD + ∠ BOE = ∠ BAC + ∠ ACE + ∠ AOC,$
$∠ BOE = ∠ AOC,$
$\therefore ∠ BPC = ∠ BAC = 90°,$
$\therefore BD ⊥ CE.$
2.如图，$△ ACB$和$△ DCE$均为等边三角形，点A，D，E在同一直线上，连接BE.求证:$AD=BE$.
答案：2.$\because △ ACB$和$△ DCE$均为等边三角形，
$\therefore AC = BC, CD = CE, ∠ ACB = ∠ DCE = ∠ CED = ∠ CDE = 60°, \therefore ∠ ACD = ∠ BCE.$
在$△ ACD$和$△ BCE$中,$\begin{cases} AC=BC, \\ ∠ ACD=∠ BCE, \\ CD=CE, \end{cases}$
$\therefore △ ACD ≌ △ BCE (\mathrm{SAS}), \therefore AD=BE.$
$\therefore AC = BC, CD = CE, ∠ ACB = ∠ DCE = ∠ CED = ∠ CDE = 60°, \therefore ∠ ACD = ∠ BCE.$
在$△ ACD$和$△ BCE$中,$\begin{cases} AC=BC, \\ ∠ ACD=∠ BCE, \\ CD=CE, \end{cases}$
$\therefore △ ACD ≌ △ BCE (\mathrm{SAS}), \therefore AD=BE.$
3. 如图,在$Rt△ ABC$中,$AB=AC$,$∠ ABC=∠ ACB=45°$,$D,E$是斜边$BC$上两点,且$∠ DAE=45°$,若$BD=3$,$CE=4$,$S_{△ ADE}=15$,则$△ ABD$与$△ AEC$的面积之和为(
A.36
B.21
C.30
D.22
B
).A.36
B.21
C.30
D.22
答案：
3.B [解析]如图,将$△ ADE$关于$AE$对称得到$△ AFE$,连接$CF$.
则$AF=AD, ∠ EAF=45°, S_{△ AFE}=S_{△ ADE}=15,$
$\therefore ∠ CAF + ∠ CAD = ∠ DAE + ∠ EAF = 45° + 45° = 90°.$
$\because ∠ BAD + ∠ CAD = ∠ BAC = 180° - ∠ ABC - ∠ ACB = 90°, \therefore ∠ CAF = ∠ BAD.$
在$△ ACF$和$△ ABD$中,$\begin{cases} AC=AB, \\ ∠ CAF=∠ BAD, \\ AF=AD, \end{cases}$
$\therefore △ ACF ≌ △ ABD (\mathrm{SAS}),$
$\therefore CF=BD=3, ∠ ACF = ∠ ABD = 45°, S_{△ ACF}=S_{△ ABD},$
$\therefore ∠ ECF = ∠ ACB + ∠ ACF = 90°,$即$△ CEF$是直角三角形,$\therefore S_{△ CEF}=\frac{1}{2}CE · CF=\frac{1}{2} × 4 × 3=6,$
$\therefore S_{△ ABD} + S_{△ AEC} = S_{△ ACF} + S_{△ AEC} = S_{△ AFE} + S_{△ CEF} = 15 + 6 = 21,$即$△ ABD$与$△ AEC$的面积之和为21.故选B.
3.B [解析]如图,将$△ ADE$关于$AE$对称得到$△ AFE$,连接$CF$.
则$AF=AD, ∠ EAF=45°, S_{△ AFE}=S_{△ ADE}=15,$
$\therefore ∠ CAF + ∠ CAD = ∠ DAE + ∠ EAF = 45° + 45° = 90°.$
$\because ∠ BAD + ∠ CAD = ∠ BAC = 180° - ∠ ABC - ∠ ACB = 90°, \therefore ∠ CAF = ∠ BAD.$
在$△ ACF$和$△ ABD$中,$\begin{cases} AC=AB, \\ ∠ CAF=∠ BAD, \\ AF=AD, \end{cases}$
$\therefore △ ACF ≌ △ ABD (\mathrm{SAS}),$
$\therefore CF=BD=3, ∠ ACF = ∠ ABD = 45°, S_{△ ACF}=S_{△ ABD},$
$\therefore ∠ ECF = ∠ ACB + ∠ ACF = 90°,$即$△ CEF$是直角三角形,$\therefore S_{△ CEF}=\frac{1}{2}CE · CF=\frac{1}{2} × 4 × 3=6,$
$\therefore S_{△ ABD} + S_{△ AEC} = S_{△ ACF} + S_{△ AEC} = S_{△ AFE} + S_{△ CEF} = 15 + 6 = 21,$即$△ ABD$与$△ AEC$的面积之和为21.故选B.
4.如图,在正方形ABCD中,E,F分别为线段BC,DC上的点,∠EAF=45°,求证:EF=BE+FD.
答案：
4.如图,延长$CB$至点$H$,使$BH=DF$,连接$AH$.
在$△ ADF$和$△ ABH$中,$\begin{cases} AD=AB, \\ ∠ D=∠ ABH=90°, \\ DF=BH, \end{cases}$
$\therefore △ ADF ≌ △ ABH (\mathrm{SAS}),$
$\therefore AF=AH, ∠ DAF = ∠ BAH.$
又$∠ DAF + ∠ BAE = 90° - 45° = 45°,$
$\therefore ∠ BAH + ∠ BAE = 45°,$即$∠ EAH = 45°.$
在$△ AEF$和$△ AEH$中,$\begin{cases} AF=AH, \\ ∠ FAE=∠ HAE=45°, \\ AE=AE, \end{cases}$
$\therefore △ AEF ≌ △ AEH (\mathrm{SAS}),$
$\therefore EF=EH=BE+BH=BE+DF.$
4.如图,延长$CB$至点$H$,使$BH=DF$,连接$AH$.
在$△ ADF$和$△ ABH$中,$\begin{cases} AD=AB, \\ ∠ D=∠ ABH=90°, \\ DF=BH, \end{cases}$
$\therefore △ ADF ≌ △ ABH (\mathrm{SAS}),$
$\therefore AF=AH, ∠ DAF = ∠ BAH.$
又$∠ DAF + ∠ BAE = 90° - 45° = 45°,$
$\therefore ∠ BAH + ∠ BAE = 45°,$即$∠ EAH = 45°.$
在$△ AEF$和$△ AEH$中,$\begin{cases} AF=AH, \\ ∠ FAE=∠ HAE=45°, \\ AE=AE, \end{cases}$
$\therefore △ AEF ≌ △ AEH (\mathrm{SAS}),$
$\therefore EF=EH=BE+BH=BE+DF.$